I am studying Kadanoff & Baym's book Quantum Statistical Mechanics and I am stuck a one point.
The are considering a system of non-interacting particles, (let's say fermions to not having to write both signs), and are then considering the G-lesser function:
G<(1,1′)=i⟨ψ†(1′)ψ(1)⟩,
where 1=r1,t1 and similarly for 1′.
Since the Hamiltonian has rotational and translational symmetry they argue that the Green's function above only depends on |r1−r1′|. Also, since the Hamiltonian is time independent the Green's function should only depend on the time difference t1−t′1. All this seems fine and I think I have sucessfully convinced myself of these facts by considering e.g. the translation operator.
However, they then define the Fourier transform as
G<(p,ω)=−i∫d3r∫dte−ip⋅r+iωtG<(r,t),
where we now use r=r1−r2 and similarly for t. Now come the claim that I cannot really see. They say that, due to the invariances I talked about above, we have
G<(p,ω)=∫dteiωtV⟨ψ†(p,0)ψ(p,t)⟩
where V is the volume of the system. Can someone please explain how this follows from the above? If I naively try to calculate this I instead get
G<(p,ω)=−i∫d3r∫dte−ip⋅r+iωtG<(r,t)=−i∫d3r∫dte−ip⋅r+iωt⟨ψ†(0,0)ψ(r,t)⟩,
which would only give the Fourier transform of the annihilation operator.
I have tried to redo everything carefully from scratch and I have found for example the relation
G<(p,ω,p′,ω′)=2πVδ(ω−ω′)δp,p′G<(p,ω).
However, I cannot really get the factor 1/V which was obtained in the book.